This is what I was given:

$5$. Find an SVD for $A=\begin{bmatrix}-2&1\\-1&0\\0&0\\1&2\end{bmatrix}$

$\text{Solution:} \ A^TA=\begin{bmatrix}-2&-1&0&1\\1&0&0&2\end{bmatrix}\begin{bmatrix}-2&1\\-1&0\\0&0\\1&2\end{bmatrix}=\begin{bmatrix}6&0\\0&5\end{bmatrix}$.
$U=\begin{bmatrix}-\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{5}}&0&-\frac{2}{\sqrt{30}}\\ -\frac{1}{\sqrt{6}}&0&0&\frac{5}{\sqrt{30}}\\ 0&0&1&0\\ \frac{1}{\sqrt{6}}&\frac{2}{\sqrt{5}}&0&\frac{1}{\sqrt{30}}\end{bmatrix}$.

I am not sure how the solutions derived the third and last columns of $U$. I was told that you could just find the null space of $A^T$. Why does that work? Could someone please give me step-by-step description of how to obtain the columns of $U$? Thank you so much.


In this problem $U$ is a $4 \times 4$ orthogonal matrix, so its columns are an orthonormal basis for $\mathbb{R}^4$. Since $A$ has rank $2$, the first two columns of $U$ are an orthonormal basis for $\DeclareMathOperator{\Col}{Col} \Col(A)$ (which is a $2$-dimensional subspace of $\mathbb{R}^4$), so we need to extend this to an orthonormal basis for $\mathbb{R}^4$. Since $\mathbb{R}^4 = \Col(A) \boxplus \Col(A)^\perp$, it suffices to find an orthonormal basis for $\Col(A)^\perp$.

Now $\DeclareMathOperator{\Nul}{Nul} \Col(A)^\perp = \Nul(A^T)$ by the definition of matrix multiplication in terms of dot products: $A^T x = 0$ iff $x$ is orthogonal to each of the rows of $A^T$ (which are the columns of $A$). We can compute a basis of $\Nul(A^T)$ (which will consist of two vectors) by row reduction, and then apply Gram-Schmidt to these vectors to get an orthonormal basis for $\Col(A)^\perp$.


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